What's wrong with my solution for this complex integral?

Question

Here I got a complex integral given by$$\frac{1}{2\pi}\int_{0}^{2\pi}{\frac{1}{z_0+2a\cos{k}}dk}$$where $z_0$ is a arbitrary complex number and $a$ is a real number. I try to solve this by taking $z=e^{ik}$ then I can substitute $\cos k=(z+\frac{1}{z})/2$, $dk=dz/iz$ in the original integral, to obtain$$\frac{1}{2\pi i}\oint{\frac{1}{az^2+z_0z+a}dz}$$The integral contour is a unit circle centered at the origin. I can evaluate this integral to obtain$$\frac{1}{2\pi i}\oint{\frac{1}{az^2+z_0z+a}dz}=\sum_i{Res[f(\lambda_i)]}$$where $\lambda_i$ denotes the singularity which lies inside the integral contour. So the integral results in$$\frac{\Theta(1-|\lambda_1|)}{a(\lambda_1-\lambda_2)}+\frac{\Theta(1-|\lambda_2|)}{a(\lambda_2-\lambda_1)}$$ where $\lambda_{1,2}=\frac{-z_0\pm\sqrt{z_0^2-4a^2}}{2a}$ and $\Theta$ denotes the HeavisideTheta function. In my solution, the result of this integral will be $0$ if both of singularities lie outside the contour, but it's totally different with the result from MMA( which couldn't be $0$ ). What's wrong with my solution?

Answer 1

For simplicity I will rescale the integral as $$ \frac1{2\pi a}\int_{0}^{2\pi}\frac{dx}{c+\cos x} =\frac1{2\pi a i}\oint_{|z|=1}\frac{dz}{z^2+2cz+1} $$ with $c=\frac{z_0}{2a}$.

The integrand has poles at: $$ z_{1,2}=-c\pm\sqrt{c^2-1}. $$

In view of $z_1\cdot z_2=1$ there are two options:

1. $|z_1|=|z_2|=1$
2. $|z_1|\ne |z_2|$.

It is not hard to demonstrate that the former case is equivalent to $c\in [-1,1]$. This corresponds exactly to the case when the original integral diverges.

In the general (second) case one of the poles is inside the integration contour and the other is outside of it. Therefore the integral is non-zero.